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By M. Aizenman (Chief Editor)

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14. For that we will need the following lemma. 6. 36) where |λ| > 0 is a parameter. Proof. 36) is obtained by direct calculation. Next, we consider the transversal fields i = p. 28) is roughly H (x − y, t − τ ; λi− ). 11) in the left quarter plane with the initial point (x, t) in it. Therefore, λi (φ(y)) ≈ λi− , and the solution indeed is about H (x − y, t − τ ; λi− ). For the more complicated case y < 0 and x > 0, G i is about H (xi− − y, t − τ ; λi− ). Here we still solve the dual equation in the left quarter plane.

Liu, Y. 29) j 2 − ∂ = ρb+ (x; λ− e−λ p y H x − y, t − τ ; −λ− p) p j 2 ∂y + ∂ j2 − ρb+ (x; −λ+p ) j eλ p y H x − y, t − τ ; λ+p 2 ∂y + ∂ j2 + + = ρb+ (x; λ− eλ p y H x − y, t − τ ; λ+p ) − ρ (x; −λ ) p p b j 2 ∂y j 2 + − ∂ eλ p y H x − y, t − τ ; λ+p − e−λ p y H x − y, t − τ ; −λ− . − ρb+ (x; λ− p) p ∂ y j2 ρb− (x; λ− p) Note that x > 1/ε and |y| ≤ 1/ε imply x − y > 0. 27). 30) for j ≥ 0 and |y| ≤ 1/ε. 23). 31) for j ≥ 0, x > 1/ε and |y| ≤ 1/ε. 23), together with x − y > 0. 29) is settled. 20).

37) (iv) For 2εt < x ≤ 3εt, Mα,β (x, t; ε, µ) = O(1) (t + 1) × 1+ 2−α 2 min β √ t 3εt −x ε , β (ε−2 ), β ε 1−α 3εt − x −1 H (x, t; ε, µ) + (t +1) 2 min √ t β (ε −2 ), β ∗ + (t + 1)− 2 min{ α−1 (ε−2 ), α−1 (t)}e−ε|x|/µ . 38) (v) For x > 3εt, Mα,β (x, t; ε, µ) = O(1) t + 1) β + (t + 1)− 2 min{ α−1 1−α 2 α−1 min β β (t), ∗ (ε−2 )}e−ε|x|/µ (ε−2 ) e−ε|x−2εt|/µ . 39) Lα,β (x, t; ε, µ) = Mα,β (−x, t; ε, µ). 6, formulated for our convenience. 7. Let µ > 0 and γ2 > γ1 > 0 be constants and λ be a parameter satisfying γ1 ε ≤ λ ≤ γ2 ε.

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